The value of $c$ for Lagrange's Mean Value Theorem for $f(x) = \sqrt{25-x^2}$ on the interval $[1, 5]$ is

In the Mean Value Theorem,$f'(c) = \frac{f(b) - f(a)}{b - a}$. If $a = 0$,$b = \frac{1}{2}$,and $f(x) = x(x - 1)(x - 2)$,then the value of $c$ is:

Let $a > 0$ and $f$ be continuous in $[-a, a]$. Suppose that $f'(x)$ exists and $f'(x) \le 1$ for all $x \in (-a, a)$. If $f(a) = a$ and $f(-a) = -a$,then $f(0)$ is:

Let $f(x)$ satisfy the requirements of Lagrange's Mean Value Theorem in $[0, 2]$. If $f(0) = 0$ and $|f'(x)| \leqslant \frac{1}{2}$ for all $x \in [0, 2]$,then-

If $a, b, c \in \mathbb{R}$ and satisfy $3a + 5b + 15c = 0$,then the equation $ax^4 + bx^2 + c = 0$ has:

Let $a, b, c$ be real numbers such that $2a + 3b + 6c = 0$ and $g(x) = ax^2 + bx + c = 0$ has at least one root in the interval $(1, 2)$. If a function $f: [1, 2] \rightarrow \mathbb{R}$ for which Rolle's Theorem holds is such that $f(x)$ is a primitive of $g(x)$,then $f(x) = $